What is the range of (4x^2+4)/(x^2+6x+9) ?

The range of (4x^2+4)/(x^2+6x+9) is f(x)>= 2/5

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range of (4x^2+4)/(x^2+6x+9)

range

\frac{4 x ^{2} + 4}{x ^{2} + 6 x + 9}

f (x) \geq \frac{2}{5}

Interval Notation

[\frac{2}{5}, \infty)

Solution steps

Rewrite as

\frac{4 x ^{2} + 4}{x ^{2} + 6 x + 9} = y

Multiply both sides by

x^{2} + 6 x + 9

4 x^{2} + 4 = y (x^{2} + 6 x + 9)

The range is the set of y for which the discriminant is greater or equal to zero

Discriminant

4 x^{2} + 4 = y (x^{2} + 6 x + 9) : 160 y - 64

160 y - 64 \geq 0 : y \geq \frac{2}{5}

Check if the range interval endpoints are included

y = \frac{2}{5} :

Included

Therefore the range is

f (x) \geq \frac{2}{5}

f (x) = \frac{1}{4} x^{6} - 5 x^{2}

e^{x + 1} - 3

y = \frac{3}{4} x

f (x) = x^{2} + 14 x + 46

\frac{x + 16}{x - 4}

What is the range of (4x^2+4)/(x^2+6x+9) ?
The range of (4x^2+4)/(x^2+6x+9) is f(x)>= 2/5